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A326988
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Sum of nonpowers of 2 dividing n.
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5
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0, 0, 3, 0, 5, 9, 7, 0, 12, 15, 11, 21, 13, 21, 23, 0, 17, 36, 19, 35, 31, 33, 23, 45, 30, 39, 39, 49, 29, 69, 31, 0, 47, 51, 47, 84, 37, 57, 55, 75, 41, 93, 43, 77, 77, 69, 47, 93, 56, 90, 71, 91, 53, 117, 71, 105, 79, 87, 59, 161, 61, 93, 103, 0, 83, 141, 67, 119, 95, 141, 71, 180, 73, 111, 123, 133, 95, 165, 79, 155
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OFFSET
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1,3
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COMMENTS
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In other words: a(n) is the sum of the divisors of n that are not powers of 2.
a(n) is also the sum of odd divisors greater than 1 of n, multiplied by the sum of the divisors of n that are powers of 2.
a(n) = 0 if and only if n is a power of 2.
a(n) = n if and only if n is an odd prime.
a(n) = 3*n/2 if and only if n is an even semiprime greater than or equal to 6 (A100484).
a(n) = n + sqrt(n) if and only if n is the square of an odd prime (see A001248 without its first term). (End)
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LINKS
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FORMULA
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a(n) = Sum_{d|n, d > 1} d * (1 - [rad(d) = 2]), where rad is the squarefree kernel (A007947) and [] is the Iverson bracket, which gives 1 if the condition is true, 0 if it's false. - Wesley Ivan Hurt, Apr 29 2020
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EXAMPLE
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For n = 18 the divisors of 18 are [1, 2, 3, 6, 9, 18]. There are four divisors of 18 that are not powers of 2, they are [3, 6, 9, 18]. The sum of them is 3 + 6 + 9 + 18 = 36, so a(18) = 36.
On the other hand, the sum of odd divisors greater than 1 of 18 is 3 + 9 = 12, and the sum of the divisors of 18 that are powers of 2 is 1 + 2 = 3, then we have that 12 * 3 = 36, so a(18) = 36.
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MAPLE
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f:= n -> numtheory:-sigma(n) - 2^(1+padic:-ordp(n, 2))+1:
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MATHEMATICA
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Table[DivisorSigma[1, n] - Denominator[DivisorSigma[1, 2n]/DivisorSigma[1, n]], {n, 100}] (* Wesley Ivan Hurt, Aug 24 2019 *)
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PROG
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(Magma) sol:=[]; m:=1; for n in [1..80] do v:=Set(Divisors(n)) diff {2^k:k in [0..Floor(Log(2, n))]}; sol[m]:=&+v; m:=m+1; end for; sol; // Marius A. Burtea, Aug 24 2019
(PARI) ispp2(n) = (n==1) || (isprimepower(n, &p) && (p==2));
a(n) = sumdiv(n, d, if (!ispp2(d), d)); \\ Michel Marcus, Aug 26 2019
(Scala) def divisors(n: Int): IndexedSeq[Int] = (1 to n).filter(n % _ == 0)
(1 to 80).map(divisors(_).filter(n => n != Integer.highestOneBit(n)).sum) // Alonso del Arte, Apr 29 2020
(Python)
from sympy import divisor_sigma
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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